Science:Math Exam Resources/Courses/MATH221/April 2010/Question 01 (e)
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Question 01 (e)
Then the matrix U is an echelon form for A (you may assume this, you don't have to do the row reduction again.)
Find a basis for the null space of A
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1.) To find the basis for the null space of A, we need to find the reduced echelon form of matrix A. In this question, it is given as matrix U.
2.) To see the pivot points clearly, we can switch columns 3 and 4 to get:
3.) Because we switched columns 3 and 4, we can see that Columns 1, 2, and 3 are linearly independent. Therefore the free variables are contained in columns 4 and 5 :
column 4 =
column 5 =
4.) Then we follow these steps in order to put matrix U in the most RREF. 1.) = - 2 2.) = + 8 3.) = - 3
The resulting RREF of matrix U is :
5.) Now we solve for each x components to solve Ax=0
Remember that we switched columns 3 and 4, therefore, (column 4) is now in (column 3) and vice versa.
=-24 =-2+8 =-2 = =
6.) Therefore, a basis for the null space of A are and